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spayn [35]
3 years ago
15

As an object falls freely toward the earth, the momentum of the object-earth system (1) decreases (2) increases (3) remains the

same ...?
Physics
1 answer:
Brums [2.3K]3 years ago
4 0

The best and most correct answer among the choices provided by your question is the third choice or number 3.

<span>As an object falls freely toward the earth, the momentum of the object-earth system remains the same.</span>


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Two forces and are applied to an object whose mass is 13.3 kg. The larger force is . When both forces point due east, the object
ANEK [815]

Answer:

Explanation:

First, It's important to remember F = ma, and in this problem m = 13.3 kg

This can be reduced to a simple system of equations problem.  Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them.  So let's call them F1 and F2, with F1 arger than F2.  Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.  

Can you solve this system of equations seeing them like this, or do you need more help?

6 0
3 years ago
a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro
finlep [7]

Answer:

\theta=53.13^o

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

V_x=V_{ox}=V_ocos\theta

V_y=V_{oy}-gt=V_osin\theta-gt

x=V_{ox}t

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}

\displaystyle y_{max}=\frac{V_{oy}^2}{2g}

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

x_{max}=3y_{max}

\displaystyle \frac{2V_{ox}V_{oy}}{g}=3\frac{V_{oy}^2}{2g}

Using the formulas for V_{ox}, V_{oy}:

\displaystyle \frac{2V_{o}cos\theta V_{o}sin\theta}{g}=3\frac{V_{o}^2sin^2\theta}{2g}

Simplifying

4cos\theta sin\theta=3sin^2\theta

Dividing by sin\theta

4cos\theta=3sin\theta

Rearranging

tan\theta=\frac{4}{3}

\theta=arctan\frac{4}{3}

\theta=53.13^o

4 0
3 years ago
A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound l
DENIUS [597]

Answer:

\mathbf{\beta = 123.75 \ dB}

Explanation:

From the question, using the expression:

125 \ dB = 10 \ log (\dfrac{I}{I_o})

where;

I_o = 10^{-12} \ W/m^2

I = 10^{12.5} \times 10^{-12} \ W/m^2

I = 3.162 \ W/m^2

This is a combined intensity of 4 speakers.

Thus, the intensity of 3 speakers = \dfrac{3.162\times 3}{4}

= 2.372 W/m²

Thus;

\beta = 10 \  log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2

\mathbf{\beta = 123.75 \ dB}

7 0
3 years ago
Identical spheres are dropped from a height of 100 m above the surfaces of Planet X and Planet Y. The speed of the spheres as a
Anna007 [38]

Answer:B

Explanation:

8 0
3 years ago
What was significant about the discovery of gallium
Kryger [21]
It confirmed medeleeve's hypothesis (prediction) and showed the use of his table
8 0
3 years ago
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